chemia - Studia

A. R. ASHRAFI, MODJTABA GHORBANI
Proof — To prove this lemma, we first notice that p and q must be
even. Consider the vertices u ij and u rs of the molecular graph of a polyhex
nanotori T = T[p,q], Figure 2. Suppose both of i and r are odd or even and σ
is a horizontal symmetry plane which maps u it to u rt , 1 ≤ t ≤ p and π is a
vertical symmetry which maps u tj to u ts , 1 ≤ t ≤ q. Then σ and π are
automorphisms of T and we have πσ(u ij ) = π(u rj ) = u rs . Thus u ij and u rs are in
the same orbit under the action of Aut(G) on V(G). On the other hand, the
map θ defined by θ(u ij ) = θ(u(p+1-i)j) is a graph automorphism of T and so if
“i is odd and r is even” or “i is even and r is odd” then again u ij and u rs will
be in the same orbit of Aut(G), proving the lemma.
▲
Theorem 3 — ξ(T[p,q]) = 3pq 2 .
Proof — From Figure 2, it can easily seen that |V(T[p,q])| = pq. By
Lemma 2, T[p,q] is vertex transitive and by Lemma 1, ξ(T[p,q]) = 3pqε(x),
for a vertex x. Now the proof is follows from this fact that ε(x) = q, proving
the result.
▲
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